Question: Simplify the following expression and state the conditions under which the simplification is valid. You can assume that $y \neq 0$. $n = \dfrac{y + 9}{-6y + 18} \div \dfrac{y + 9}{y^2 + y - 12} $
Solution: Dividing by an expression is the same as multiplying by its inverse. $n = \dfrac{y + 9}{-6y + 18} \times \dfrac{y^2 + y - 12}{y + 9} $ First factor the quadratic. $n = \dfrac{y + 9}{-6y + 18} \times \dfrac{(y - 3)(y + 4)}{y + 9} $ Then factor out any other terms. $n = \dfrac{y + 9}{-6(y - 3)} \times \dfrac{(y - 3)(y + 4)}{y + 9} $ Then multiply the two numerators and multiply the two denominators. $n = \dfrac{ (y + 9) \times (y - 3)(y + 4) } { -6(y - 3) \times (y + 9) } $ $n = \dfrac{ (y + 9)(y - 3)(y + 4)}{ -6(y - 3)(y + 9)} $ Notice that $(y + 9)$ and $(y - 3)$ appear in both the numerator and denominator so we can cancel them. $n = \dfrac{ \cancel{(y + 9)}(y - 3)(y + 4)}{ -6\cancel{(y - 3)}(y + 9)} $ We are dividing by $y - 3$ , so $y - 3 \neq 0$ Therefore, $y \neq 3$ $n = \dfrac{ \cancel{(y + 9)}\cancel{(y - 3)}(y + 4)}{ -6\cancel{(y - 3)}\cancel{(y + 9)}} $ We are dividing by $y + 9$ , so $y + 9 \neq 0$ Therefore, $y \neq -9$ $n = \dfrac{y + 4}{-6} $ $n = \dfrac{-(y + 4)}{6} ; \space y \neq 3 ; \space y \neq -9 $